Question 312439
Find a three digit number such that if the digits at the tens place and the hundreds place are reversed then the number obtained is twenty percent greater than the original number.

b*100 + a*10 + c = 1.2 * (a*100 + b*10 + c)
100b + 10a + c = 120a + 12b + 1.2c
10a - 120a + 100b - 12b + c - 1.2c = 0
-110a + 88b - 0.2c = 0
110a - 88b + 0.2c = 0 (divided by -1)
set c = 0
try 9 -> 4+5=9, 5+4=9
110(4) - 88(5) = 440 - 440 = 0
x = 1.2 * 450
x = 540
the 3 digit number is 450